Constructing Isomorphism between finite field

One way to do it is just to slug through it:

An isomorphism $\mathbb{F}_3(\alpha)\to\mathbb{F}_3(\beta)$ is completely determined by the image of $\alpha$; and, as you note in comments, the image of $\alpha$ must be of the form $a\beta^2 + b\beta + c$, with $a,b,c\in\mathbb{F}_3$. Moreover, the image of $\alpha$ must satisfy $\alpha^3 = \alpha - 1$; that is, you want to find $a,b,c$ such that $$(a\beta^2 + b\beta + c)^3 = a\beta^2 + b\beta + c-1.$$ So you can just expand the left hand side, using $\beta^3 = \beta^2-1$, and figure out the coefficients. We have: $$\begin{align*} \beta^3 &= \beta^2 - 1\\ \beta^4 &= \beta^3-\beta\\ &= \beta^2 - \beta - 1\\ \beta^6 &= (\beta^3)^2 = \beta^4 - 2\beta^2 + 1\\ &= \beta^2 - \beta - 1 -2\beta^2 + 1\\ &= -\beta^2 - \beta. \end{align*}$$ And so, since we are in characteristic $3$, $$\begin{align*} (a\beta^2 + b\beta + c)^3 &= a^3\beta^6 + b^3\beta^3 + c^3\\ &= a^3(-\beta^2-\beta) + b^3(\beta^2-1) + c^3\\ &= (b^3-a^3)\beta^2 + (-a^3)\beta + c^3-b^3\\ &= a\beta^2 + b\beta + c - 1. \end{align*}$$ So we need to solve the equations $$\begin{align*} b^3-a^3 &= a\\ -a^3 &= b\\ c^3 -b^3&= c-1. \end{align*}$$ Since $a,b,c\in\mathbb{F}_3$, where $x^3=x$ for all $x$, we get $$\begin{align*} b-a &= a\\ -a&=b\\ c-b&= c-1 \end{align*}$$ The first two equations both give $b=-a$; the last equation gives $b=1$. So $a=-1$, $b=1$, and $c$ is free (this gives the three roots of the polynomial). That is, $f(\alpha)$ can be any of $-\beta^2+\beta$, $-\beta^2+\beta+1$, $-\beta^2+\beta-1$. You can verify they all work.

From $\beta^3 - \beta^2 + 1 = 0$, we get $1 - \beta^{-1} + \beta^{-3} = 0$, and since $\alpha^3 - \alpha + 1 = 0$, we can map $\alpha$ to $\beta^{-1}$ to get the desired isomorphism. Of course, from $\beta^3 - \beta^2 + 1 = 0$ we get $\beta^2 - \beta + \beta^{-1} = 0$, that is, $\beta^{-1} = -\beta^2 + \beta$. In other words, we can take $f(\alpha) = -\beta^2 + \beta$ exactly as Arturo Magidin found.

Edit: Added note We could instead have chosen to map $\alpha$ to the conjugates $\beta^{-3}$ or $\beta^{-3^2}$ of $\beta^{-1}$ where we have that $$\beta^{-3} = \beta^{-1} - 1 = -\beta^2 + \beta - 1,$$ and $$\beta^{-3^2} = (\beta^{-1} - 1)^3 = \beta^{-3}-1 = -\beta^2 + \beta + 1$$which are the other two possible images of $\alpha$ given in Arturo's answer. As in implied in Jyrki Lahtonen's comment, a simpler derivation is possible in this case because $\alpha$ and $\beta$ are roots of reciprocal polynomials.